Transcript PPT

THE FIRST LAW of THERMODYNAMICS: Conservation of energy
Heat input dq
Work done BY system dw
If we identify du as the change in internal energy of the system (associated with
changes in the kinetic and potential energy of the molecules),we write
The “First Law” of Thermodynamics as:
dq = du + dw
If the work is mechanical work (expansion work), as we defined last time, then we
can write dw = pdv, and
dq = du + pdv
Internal energy is a STATE VARIABLE: the change in internal energy is the
difference between initial and final states. du is independent of path between
initial and final states.
The state variables, like u, are functions of any TWO independent intensive
variables, e.g., p, T, v (specific volume). E.g., we can express u = u(v,T).
JOULE’S LAW: Joule’s experiment (a thought experiment here) demonstrated that,
for an Ideal Gas, the internal energy u is a function of T only. Let’s see how to
show this:
Consider a gas expanding into an evacuated cylinder (vacuum). Since p is zero, no
mechanical work is done and dw = 0.
Imagine that the process is also adiabatic (perfectly insulated walls), so dq = 0.
Since dq = dw = 0, du = 0.
Under these conditions, its clear that the volume of the gas changed, as did the
potential energy of the molecular configuration. Therefore the internal energy of the
gas must be independent of volume, and is only a function of T.
Joule showed experimentally that the temperature of the gas in such an experiment
remained constant.
Joule’s Law states: “if a gas expands without doing any work, and under adiabatic
conditions, the temperature of the gas remains constant.”
Corollary: Molecules of an ideal gas do not exert any attractive or repulsive forces
on one another.
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Specific Heats and Heat Capacities
Specific heat and heat capacities for various substances allow for T changes to be
calculated per unit of heat added to the substance.
If heat is added to a substance at constant pressure, we define the specific heat at
constant pressure, cp, as
 
q
c p   
T p
(aside: m Cp is the HEAT CAPACITY – unit difference)
If volume remains constant during heating, we define the specific heat at constant
volume, cv, as

q 
cv   
T v
If we know what values these properties have for ideal gases, then we can readily
compute temperature changes that occur under addition of heat.

We’ll start by deriving
equivalent forms of the First Law for ideal gases that are
easier to work with.
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Start with the definition of cv:
q 
c v   
T v
But if dv = 0 then dw = 0, and the First Law reduces to dq = du
Therefore
u 
c v   
T v
If we consider an Ideal Gas, u depends only upon T.
Therefore the restriction on constant volume can be
dropped, and
du
(generally true)
c 

v
dT
Use p  RT
(Please note v and α
denote same thing)
dq  cv dT  pd
Express the last term differently: dq  cv dT  d ( p )   dp
 cv dT  RdT   dp
 (cv  R)dT   dp
At constant pressure,  dp  0 and equation becomes dq  (cv  R)dT
q 
, we see c p  cv  R
Since c p   
dq  c p dT   dp
T p
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Equivalent
 form of first law for an ideal gas
First Law can then be written as
Since
c p  cv  R; c p  cv
In a constant pressure process, volume expands, and hence some of the heat
added goes into expanding the volume instead of increasing T
For dry air;
cv  717 J deg 1 kg 1
c p  717  287  1004 J deg 1 kg 1
dq
dq
dT 
or
cv
cp
Where did those values come from? (let’s see how cp and cv values are derived).

1
1
From statistical mechanics, u  kT  RT which is an expression for the
2
2
per molecule
per mole
Internal Energy per degree of freedom. (This expression is valid as far as kT <<
electronic excitation levels (<104 °K)).
For monatomic gases we have 3 degrees of freedom associated with x,y,z direction
velocities. Hence
3
u  RT Hence c  3 R (per unit mass), since c  du
v
v
2
2
Also, since
c p  cv  R
we have c p 
5
R
2
dT
5
For diatomic molecules, more closely representative of air (e.g. O2, N2), we
have higher order degrees of freedom. Consider the following model,
x
y
a diatomic molecule is oriented
along the y-axis
z
- This molecule has 3 degrees of freedom for translational motion and 2
rotational modes (x,z axes)
-The moments of inertia around the x & z axes are very much larger than the
moments of inertia around the y-axis.
- The molecule also has 2 vibrational modes (position & velocity) but the
vibrational modes are usually dominated by the rotational modes for
atmospheric (troposphere/stratosphere) temperatures. Thus,
u
5
RT so
2
7
5
and
c

R
cv  R
p
2
2
Therefore for dry air (modeled as an ideal, diatomic gas) we can estimate
5
5
cv  Rd  (287.05 J deg 1 kg 1 )  717 J deg 1 kg 1
2
2
7
c p  Rd  1004 J deg 1 kg 1
2
(Note units)
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Quick Summary thus far:
• The First Law is du = dq – dw
• Heat capacities are defined as partial derivatives of q with respect to T, at
constant v OR p.
• The internal energy of an ideal gas is a function of temperature only, so
the following is always true:
du  cv dT
• For an ideal gas, we can then write the First Law as either

dq  cv dT  pdv
dq  c p dT  vdp
 be found theoretically for monatomic and diatomic gases
• cp and cv can
• So we can compute T changes (and thus internal energy changes, since

they depend only on T!) if we know the heat in/out and dv or dp changes
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ADIABATIC PROCESSES
An adiabatic process involves no heat exchange between a substance & its
surroundings. Hence dq = 0
Therefore, for an adiabatic transformation, dw = -du = cv dT
 Work done is equal to the decrease in internal energy
In this example
two different paths
are used to go
from V1 to V2.
PV = n RT
http://en.wikipedia.org/wiki/Adiabatic_process
Notice that the adiabat is drawn steeper than the isotherm. Why?
During the adiabatic compression dq = 0 and p dv <0 (volume decreases)
To obey the First Law (du=dw), T must increase. In contrast, T remains
constant during the isothermal compression. Since the final temperature is
higher in the adiabatic compression, therefore (by Ideal Gas Law), final P must
also be higher than for isothermal process: So adiabat is steeper than isotherm.
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The Adiabatic Air Parcel
Idealized model of an air parcel is a small volume of air:
• Thermally insulated from the environment so temperature changes are adiabatic
• Parcel pressure immediately adjusts to environmental pressure
• Parcel moves slowly enough such that macroscopic KE is negligible portion of total energy
Consider such a parcel moving about under adiabatic conditions
What is the change in temperature with height?
Use First Law and assume Ideal Gas:
0
or
 c p dT 
1
dp

c p dT   gdz
dq  du  dw
0  c p dT   dp
We chose this form over
dq  cv dT  pd
using the hydrostatic equation
dT  g

 d
dz c p
d  9.8C / km
dp
  g
dz
Dry adiabatic
lapse rate
In the above derivation we used the hydrostatic equation
-Derivation
p  dp  p   gdz
p
p  dp
(force balance)
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The Potential Temperature: Of fundamental importance to adiabatic
processes in the atmosphere is the concept of potential temperature, q
Again use the First Law for an adiabatic process,
0  c p dT 
RT
dp
p
c p dT dp

0
R T

or
Integrating from
Po
cp
R
where
0  c p dT   dp
p  RT
T  q to some point P, T
qT d ln T   PP d ln P
P
P

which yields, q  T ( )
R / cp
Poisson’s Equation
q is the temperature a parcel would acquire if it ascended or descended
adiabatically from pressure P to P
(e.g., choose P  1000mb )
1
1
R 287 J deg kg

 0.286
1
1
c p 1004 J deg kg
We expect potential temperature to be conserved for (dry) adiabatic
transformations. As the parcel moves about in the atmosphere under
adiabatic conditions, its potential temperature q does not change
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Using potential temperature
• For an adiabatic process,
c P dT  dP  RT
dP
P
T  P k
k
T  
P
    and   

q  1000 mbar 
T0  P0 
k
R (c P  c v )

 0.286
cP
cP
• θ is the temperature that a parcel of air would
have if, starting with temperature T and pressure
P, it were moved to a final pressure of 1000 mbar

• θ is conserved in an adiabatic process
• Lines of constant θ are also lines of constant
entropy (isentropes).
• Stüve diagram (1927): Notice from eqn, for a
chosen θ, T varies as Pk
• Constant-θ lines are straight (solid) lines if
we choose the axes properly
• Which are isobars, isotherms?
http://www.vcrlter.virginia.edu/~bph/AW_Book_Spring_96/AW_Book_41.html
Pk
T
Extension: atmospheric stability
We already saw that the dry adiabatic lapse rate is
dT
g
     0.98C/100m
dz
cP
What is the buoyant force on a parcel of air? Net buoyant force upward is Vg()

dry air
V, T, ρ
Displaces ambient air
V, T’, ρ’
ρgV
’gV
Acceleration of the parcel is equal to the sum of the forces -- here, only bouyancy, so:
   T  T 
d 2z
 g

 g
2


dt
T 
  
If parcel is warmer than ambient air (T > T’),
force is positive and parcel moves upward
A dry parcel that starts at temperature T and is lifted Δz has a new temperature T – Δz
Let’s suppose the environmental T changes with height at a different rate, so the ambient

temperature is T – gΔz. The excess temperature of the parcel over ambient air is Δz (g )
Atmospheric stability, continued
When Δz (g ) is positive, the parcel is warmer than its surroundings and
is accelerated upward (unstable atmosphere); if the opposite, a restoring
force acts against the direction of any motion (stable atmosphere); if the
atmosphere is neutral, its lapse rate is the dry adiabatic lapse rate
In terms of q
T
kT
dT  dq 
dP
q
P
1 q 1 T k P 1


   g 
q z T z P z T

q
0
z
q
0
z
q
0
z
stable
Pk
neutral
unstable
T
Other State Variables: Enthalpy
Consider the addition of heat to a substance at constant pressure so that the
specific volume increases from v1 to v2 ; the work done is p(v2 – v1 )
v1
v2
From First Law, dq  (u  u )  p(v  v )
2
1
2
1
 (u2  pv2 )  (u1  pv1 )
where h = u + pv
Still assuming constant p so
work can be written this way
Specific quantities
are used here
 h2  h1
 definition of ENTHALPY
For constant pressure process, dq  dh  h2  h1 that is, the change
in enthalpy
is equal to the heat added (or heat extracted).
Complete derivative ( is the same as v above)
dh  du  d ( p )
dq  c p dT   dp
 du  pd   dp
 dq   dp
(du  dq  pd )
 dp  0
 dq when dp  0
Units of h  J kg 1 or
J mole
1
dq  dh  c p dT
Enthalpy is useful for studying processes at constant pressure, and open systems.
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Since h = u + Pv and, for an ideal gas, Pv = RT, then for the ideal gas h = u + RT
R is constant, and u = u(T) only, so we conclude h = h(T) only for an ideal gas.
Aside: application of enthalpy balance in steady-flow system:
energy balance for a single-stream, steady-flow system with negligible changes in kinetic
and potential energies: q – ws = Dh, where ws is shaft work
In the system shown (flow through a “throttling valve”), the pressure drops across the
valve, but no shaft work is done. The flow through the valve is also adiabatic, since there
is not much time or area for heat transfer. So we conclude Dh = 0 and such a valve is an
“isenthalpic device” (internal and flow energies are interconverted).
From h = h(T,P), taking total derivative

v  
h 
h 
dh  0  c p dT  v  T  dP
dh

dT



  dP and using a Maxwell relation
T P 

T P
P T
v   T 
1 
 v  T     “Joule-Thompson coefficient”, = 0 for an ideal gas
T P  P h (show…)
c p 

The J-T effect is applied in the Linde technique, where the cooling effect is used to liquefy gases, and
also in many cryogenic applications (e.g. for the production of liquid oxygen, nitrogen, and argon).

The Second Law of Thermodynamics
(http://en.wikipedia.org/wiki/Second_law_of_thermodynamics)
Clausius’ version:
No process is possible whose sole result is the transfer of heat from a body
of lower temperature to a body of higher temperature
The second law of thermodynamics is an expression of the tendency that
over time, differences in temperature, pressure, and chemical potential
equilibrate in an isolated physical system. From the state of
thermodynamic equilibrium, the law deduced the principle of the increase of
entropy and explains the phenomenon of irreversibility in nature. The second
law declares the impossibility of machines that generate usable energy from
the abundant internal energy of nature by processes called perpetual motion
of the second kind.
In classical thermodynamics, the second law is a basic postulate applicable
to any system involving measurable heat transfer, and defines the concept of
thermodynamic entropy.
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The Second Law and Cyclic Processes
The Second Law deals with the maximum amount, or fraction of, heat that can be
converted to “work”. For any thermodynamic system there is a theoretical limit to the
conversion factor between heat & work.
To study this process further – consider the CARNOT CYCLE (a cyclic process)
CYCLIC PROCESS – a series of change in the state of a substance in which its
volume changes and it does work, with the substance returning to its original state.
Initial & final state are identical in a cyclic process.
Since the initial & final states are the same, du  0 for a cyclic process.
Therefore dq  dw for a cyclic process.
This implies
Net work done in a cyclic process is equal to the heat
added to the system.
Consider the following “heat engine”
Input Heat Q1
Engine
The “efficiency” of this heat engine

Q
work done Q1  Q2

 1 2
heatinput
Q1
Q1
Q2

If
Exhaust Heat
is defined as
Q2  0
then obviously 
1
A “perfect” engine
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CARNOT CYCLE
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